Question

A ball is dropped, from

rest, from the top of a building. There are two motion sensors positioned

outside of two windows in the building to record the velocities of the ball

as it passes them. Sensor A is located above sensor B. Sensor A records

the ball’s velocity to be -15.5 m

s

and sensor B records the ball’s velocity to

be -17.2 m

s

. a) how far apart are the sensors?
b) how far from sensor a is the top of the building where the ball was released
c) if the time from when the ball passes sensor b until it hits the ground is 3.5s how tall is the building

Answer 1

The sensors are approximately
`2.83\; {\rm m}` apart.

Sensor
`\texttt{a}` is approximately
`12.2\; {\rm m}` from the top of the building.

Height of the building is approximately
`135\; {\rm m}`.

(Assumptions: air resistance on the ball is negligible;
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Make use of the SUVAT equation
`x = (v^(2) - u^(2)) / (2\, a)`, where
`x` represents displacement,
`v` and
`u` are the final and initial velocity, and
`a` is the acceleration.

In other words, as the velocity of the object changes from
`u\!` to
`v\!` at a rate of
`a`, position of the object would have changed by
`x = (v^(2) - u^(2)) / (2\, a)`.

a)

When the ball is at sensor
`\texttt{a}`, velocity of the ball was
`u = (-15.5)\; {\rm m\cdot s^(-1)}`.

Shortly after, when the ball is at sensor
`\texttt{b}`, velocity of the ball was
`v = (-17.2)\; {\rm m\cdot s^(-2)}`.

Under the assumptions,
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`. Apply the SUVAT equation
`x = (v^(2) - u^(2)) / (2\, a)` to find the change in the position of the ball between sensor
`\texttt{a}` (
`u = (-15.5)\; {\rm m\cdot s^(-1)}`) and
`\texttt{b}` (
`v = (-17.2)\; {\rm m\cdot s^(-2)}`):

`\begin{aligned}x &= ((-17.2)^(2) - (-15.5)^(2))/(2\, (-9.81)) \; {\rm m} \approx (-2.83)\; {\rm m}\end{aligned}`.

Hence, the distance between sensor
`\texttt{a}` and
`\texttt{b}` is approximately
`2.83\; {\rm m}`.

b)

Since the ball was released from rest, the initial velocity of the ball would be
`u = 0\; {\rm m\cdot s^(-1)}`.

Let
`v` denote the velocity of the ball at sensor
`\texttt{a}`:
`v = (-15.5)\; {\rm m\cdot s^(-1)}`.

Apply the SUVAT equation
`x = (v^(2) - u^(2)) / (2\, a)` to find the change in the position of the ball between the top of the roof (
`u = 0\; {\rm m\cdot s^(-1)}`) and sensor
`\texttt{a}` (
`v = (-15.5)\; {\rm m\cdot s^(-1)}`):

`\begin{aligned}x &= ((-15.5)^(2) - (0)^(2))/(2\, (-9.81)) \; {\rm m} \approx (-12.2)\; {\rm m}\end{aligned}`.

In other words, the distance between sensor
`\texttt{a}` and the top of the roof would be approximately
`12.2\; {\rm m}`.

c)

Another SUVAT equation,
`v = u + a\, t`, gives the velocity of an object after accelerating for a duration of
`t` (at a rate of
`a`, starting from an initial velocity of
`u`.)

Make use of this SUVAT equation to find the velocity of the ball right before hitting the ground. Specifically, velocity of the ball was
`u = (-17.2)\; {\rm m\cdot s^(-1)}` at sensor
`\texttt{b}`. After accelerating at
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` for
`t = 3.5\; {\rm s}`, velocity of the ball would be:

`\begin{aligned}v &= u + a\, t \\ &= (-17.2)\; {\rm m\cdot s^(-1)} + (-9.81)\, (3.5)\; {\rm m\cdot s^(-1)} \\ &= (-51.535)\; {\rm m\cdot s^(-1)} \end{aligned}`.

Again, the velocity of the ball was
`u = 0\; {\rm m\cdot s^(-1)}` at the top of the building. Apply the SUVAT equation
`x = (v^(2) - u^(2)) / (2\, a)` to find the change in the position of the ball between the top of the building (
`u = 0\; {\rm m\cdot s^(-1)}`) and right before hitting the ground (
`v = (-51.535)\; {\rm m\cdot s^(-1)}`):

`\begin{aligned}x &= ((-51.535)^(2) - (0)^(2))/(2\, (-9.81)) \; {\rm m} \approx (-135)\; {\rm m}\end{aligned}`.